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3 years ago

What happens in double transverse wishbone front suspension when brakes are applied.

Engineering

1 answer:

RideAnS [48]3 years ago

5 0

Answer:

When the brakes are applied the in the typical double transverse wishbone front suspension, it "drives" the car ground due to the setting of the link-type system pivot points on the lower wishbone are have parallel alignment to the road

Explanation:

In order to minimize the car's reaction to the application of the brakes, the front and rear pivot are arranged with the lower wishbone's rear pivot made to be higher than the front pivot as such the inclined wishbone torque results in an opposing vertical force to the transferred extra weight from the back due to breaking.

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Consider a fan located in a 3 ft by 3 ft square duct. Velocities at various points at the outlet are measured, and the average f

natulia [17]

Answer:

minimum electric power consumption of the fan motor is 0.1437 Btu/s

Explanation:

given data

area = 3 ft by 3 ft

air density = 0.075 lbm/ft³

to find out

minimum electric power consumption of the fan motor

solution

we know that energy balance equation that is express as

E in - E out = $\frac{dE \ system}{dt}$ ......................1

and at steady state $\frac{dE \ system}{dt}$ = 0

so we can say from equation 1

E in = E out

minimum power required is

E in = W = m $\frac{V^2}{2}$ = $\rho A V \frac{V^2}{2}$

put here value

E in = $\rho A V \frac{V^2}{2}$

E in = $0.075 *3*3* 22 \frac{22^2}{2}$

E in = 0.1437 Btu/s

minimum electric power consumption of the fan motor is 0.1437 Btu/s

5 0

3 years ago

What is the resistance of a resistor if the current flowing through it is 3mA and the voltage across it is 5.3V?

Flura [38]

Answer: 1766.667 Ω = 1.767kΩ

Explanation:

V=iR

where V is voltage in Volts (V), i is current in Amps (A), and R is resistance in Ohms(Ω).

3mA = 0.003 A

Rearranging the equation, we get

R=V/i

Now we are solving for resistance. Plug in 0.003 A and 5.3 V.

R = 5.3 / 0.003

= 1766.6667 Ω

= 1.7666667 kΩ

The 6s are repeating so round off to whichever value you need for exactness.

6 0

1 year ago

Steam enters a two-stage adiabatic turbine at 8 MPa and 5008C. It expands in the first stage to a state of 2 MPa and 3508C. Stea

Nataly [62]

Answer:

1) The exergy of destruction is approximately 456.93 kW

2) The reversible power output is approximately 5456.93 kW

Explanation:

1) The given parameters are;

P₁ = 8 MPa

T₁ = 500°C

From which we have;

s₁ = 6.727 kJ/(kg·K)

h₁ = 3399 kJ/kg

P₂ = 2 MPa

T₂ = 350°C

From which we have;

s₂ = 6.958 kJ/(kg·K)

h₂ = 3138 kJ/kg

P₃ = 2 MPa

T₃ = 500°C

From which we have;

s₃ = 7.434 kJ/(kg·K)

h₃ = 3468 kJ/kg

P₄ = 30 KPa

T₄ = 69.09 C (saturation temperature)

From which we have;

h₄ = $h_{f4}$ + x₄× $h_{fg}$ = 289.229 + 0.97*2335.32 = 2554.49 kJ/kg

s₄ = $s_{f4}$ + x₄× $s_{fg}$ = 0.94394 + 0.97*6.8235 ≈ 7.563 kJ/(kg·K)

The exergy of destruction, $\dot X_{dest}$ , is given as follows;

$\dot X_{dest}$ = T₀ × $\dot S_{gen}$ = T₀ × $\dot m$ × (s₄ + s₂ - s₁ - s₃)

$\dot X_{dest}$ = T₀ × $\dot W$ ×(s₄ + s₂ - s₁ - s₃)/(h₁ + h₃ - h₂ - h₄)

∴ $\dot X_{dest}$ = 298.15 × 5000 × (7.563 + 6.958 - 6.727 - 7.434)/(3399 + 3468 - 3138 - 2554.49) ≈ 456.93 kW

The exergy of destruction ≈ 456.93 kW

2) The reversible power output, $\dot W_{rev}$ = $\dot W_{}$ + $\dot X_{dest}$ ≈ 5000 + 456.93 kW = 5456.93 kW

The reversible power output ≈ 5456.93 kW.

6 0

3 years ago

When lining up the song on the tempo grid it is important to allow

Reika [66]

Tempo decides the speed at which the music is played.

<u>Explanation:</u>

The Tempo of a bit of music decides the speed at which it is played, and is estimated in beats per minute (BPM). The 'beat' is dictated when mark of the piece, so 100 BPM in 4/4 compares to 100 quarter notes in a single moment.

A quick tempo, prestissimo, has somewhere in the range of 200 and 208 beats for each moment, presto has 168 to 200 beats for every moment, allegro has somewhere in the range of 120 and 168 beats for every moment, moderato has 108 to 120 beats for every moment, moderately slow and even has 76 to 108, adagio has 66 to 76, larghetto has 60 to 66, and largo, the slowest rhythm, has 40 to 60.

6 0

3 years ago

There are two types of cellular phones, handheld phones (H) that you carry and mobile phones (M) that are mounted in vehicles. P

nexus9112 [7]

Answer:

A) P(W) = 0.5

B) P(MF) = 0.3

C) P(H) = 0.6

Explanation:

We are told that there are two types of cellular phones which are handheld phones (H) that you carry and mobile phones (M) that are mounted in vehicles.

Also, Phone calls can be classified by the traveling speed of the user as fast (F) or slow (W).

Thus, the sample space is combination of types and classification we are given and it is written as;

S = {HF, HW, MF, MW}

A) Now, phones can either be fast(F) or slow(W). Thus, we can write;

P(F) + P(W) = 1

We are given P(F) = 0.5

Thus;

0.5 + P(W) = 1

P(W) = 1 - 0.5

P(W) = 0.5

B) Now, from the problem statement, a phone call can either be made with a handheld(H) or mobile(M). Thus the sample space partition is {H, M} and we can express as;

P(H ∩ F) + P(M ∩ F) = P(F)

We are given P[F] = 0.5 and P[HF] = 0.2.

P(H ∩ F) is same as P[HF]

Also, P(M ∩ F) is same as P(MF)

Thus;

0.2 + P(MF) = 0.5

P(MF) = 0.5 - 0.2

P(MF) = 0.3

C) Similarly, mobile Phone calls can either be fast or slow. It means the sample space partition is {F, W}

Thus;

P(M) = P(MW) + P(MF)

P(M) = 0.1 + 0.3

P(M) = 0.4

Now, since cellular phones can either be handheld(H) or Mobile(M), then we can say;

P(H) + P(M) = 1

P(H) + 0.4 = 1

P(H) = 1 - 0.4

P(H) = 0.6

5 0

3 years ago